There are lots of cases where this assumption doesn't hold: if you're sampling from a join, or from a table with non-contiguous ID values, for example.
It seems less, not more, likely that this kind of situation would exist in a huge table - what use would such a table be if you could only process it sequentially?
A sufficiently smart database might understand "order by rand() limit n" and do reservoir sampling on row id's behind the scenes. I wouldn't count on it, though.
If you set it up in advance, you could also use reservoir sampling to avoid even storing all the data in the first place, while still saving a random sample.
In fact, i think this is the underlying meme of 'big data' - that it's now plausible and useful to do analysis over all of a large dataset, so you might as well build everything around iterating over the entire table regardless.
Unfortunately I don't understand how each element has the same probability of being chosen. I.e. the first element seems to have a probability of 1/n, while the second has 1/n + 1/(n-1) etc. Am i wrong about this?
/* This is standard modern C++ random number generation.
It means, using the random generator "engine", return
a number according to distribution "dist". In this
"dist" is a uniform distribution from "i" to "max-min".
"auto" is the new way to type variables in C++: it
infers the type of the variable from the type of the
expression */
22: auto index = dist(engine);
/* The following lines are a bit more complex, as they
contain a nested expression. Let us focus on the inner
expression:
(index < size ) ? array[index]
: map.insert({index, min + index}).first->second
This is in the form:
test-expr ? when-true-expr
: when-false-expr
This expression is a ternary condition. What it means
is that the expression before the question mark is
computed (index < size), and if it evaluates to true,
the overall expression will resolve to the result of
the expression between the question mark and the colon,
otherwise it resolves to what is after the colon.
Here, it means that if "index" is smaller than "size",
use "array[index]", otherwise insert a new entry in the
map, with key "index" and value "min+index", then reuse
the result of the operation by accessing member "second"
of the member "first". I am not familiar with STL maps;
here is an explanation of the map::insert method from
cplusplus.com:
The single element versions (1) return a pair, with
its member pair::first set to an iterator pointing
to either the newly inserted element or to the
element with an equivalent key in the map. The
pair::second element in the pair is set to true if
a new element was inserted or false if an equivalent
key already existed.
*/
23: std::swap(array[i], index < size
24: ? array[index]
25: : map.insert({index, min + index}).first->second);
/* Then, the std::swap operation simply swaps the two indicated
values in the array. */I've annotated the program here: http://ideone.com/gmaBri
> Unfortunately I don't understand how each element has the same probability of being chosen. I.e. the first element seems to have a probability of 1/n, while the second has 1/n + 1/(n-1) etc. Am i wrong about this?
This is just a partial Fisher-Yates shuffle[1] using a sparse sequence instead of an array. Keep in mind that if an element is not chosen it is swapped with the chosen element and will be considered again. I think the best way to think about it is that you are choosing a random element from the set and then "removing" it from the set with the swap and then repeating.
[1] http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle#Th...
I had the same thought reading this. This solution is only really suitable if the number of elements to select is within some constant of the total number of elements.
> More generally, reservoir sampling is appropriate when you don't know the size of the set beforehand. But in the premise, we do know the size of the set. And if you know the size of the set (and have some way to address arbitrary elements), then randomly generating a set of X indices is going to be much simpler and faster. Yeah, you have to deal with duplicate indices, but that's an easy problem to solve.
Last time I had to solve this, I came up with the following solution[1] which doesn't generate any duplicates and runs in O(k) where k is the size of result set.
[1] http://ideone.com/CAwuxW